Is it possible with Fixture to create a list of N objects?

Question

I want to create with Fixture a list of N objects.

I know I can do it with:

List<Person> persons = new List<Person>();

for (int i = 0; i < numberOfPersons; i++)
{
    Person person = fixture.Build<Person>().Create();
    persons.Add(person);
}

Is there any way I can use the CreateMany() method or some other method in order to avoid the loop?

score 7 · Answer 1 · answered Oct 10 '14 at 12:22

7

Found the answer. CreateMany has some overloads that get 'count'.

Thanks people.

answered Oct 10 '14 at 12:22

Efstathios Chatzikyriakidis

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1

how did you find the answer just within 100 seconds since you posted the question !! :P – Neel Oct 10 '14 at 12:23
When 'count' is zero it does not return an empty collection but it throws an expception. Bad implementation. – Efstathios Chatzikyriakidis Oct 10 '14 at 12:24

thumbmunkeys · Answer 2 · 2014-10-10T12:28:59.210

6

you can use linq:

  List<Person> persons = Enumerable.Range(0, numberOfPersons)
            .Select(x => fixture.Build<Person>().Create())
            .ToList();

edited Oct 10 '14 at 12:28

answered Oct 10 '14 at 12:22

thumbmunkeys

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I will accept it for now. But, I would like a solution with a more Fixture way rather than LINQ. – Efstathios Chatzikyriakidis Oct 10 '14 at 12:25
Not working. It shows a warning that Select has two overloaded methods. – Efstathios Chatzikyriakidis Oct 10 '14 at 12:28
Thanks. Is there a solution by using only Fixture? – Efstathios Chatzikyriakidis Oct 10 '14 at 12:30
Nowaday at least you can also, in a one liner, adapt any property using linq: List persons = Enumerable.Range(0, numberOfPersons) .Select(x => fixture.Build() .With(p => p.Name, "Some name") .With(p => p.Address, "Some address").Create()) .ToList(); – Henrik Aug 23 '21 at 08:03

score 4 · Answer 3 · answered Jan 23 '19 at 11:14

Yes Sure, you can use the CreateMany as the next sample:

var numberOfPersons = 10; //Or your loop length number
var fixture = new Fixture();
var person = fixture.CreateMany<Person>(numberOfPersons).ToList(); 
//ToList() to change  the IEnumerable to List

score 1 · Answer 4 · answered Jan 18 '18 at 02:05

1

var dtos = (new Fixture()).CreateMany<YourObjectType>(numberRecords);

answered Jan 18 '18 at 02:05

Wolf

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Efstathios Chatzikyriakidis · Accepted Answer · 2014-10-10T13:06:49.083

I have done it people.

/// <summary>
/// This is a class containing extension methods for AutoFixture.
/// </summary>
public static class AutoFixtureExtensions
{
    #region Extension Methods For IPostprocessComposer<T>

    public static IEnumerable<T> CreateSome<T>(this IPostprocessComposer<T> composer, int numberOfObjects)
    {
        if (numberOfObjects < 0)
        {
            throw new ArgumentException("The number of objects is negative!");
        }

        IList<T> collection = new List<T>();

        for (int i = 0; i < numberOfObjects; i++)
        {
            collection.Add(composer.Create<T>());
        }

        return collection;
    }

    #endregion
}

score 0 · Answer 6 · answered Feb 17 '22 at 17:37

0

var people = _fixture
        .Build<Person>()
        .CreateMany()
        .ToList(); // if you want to return a generic list

answered Feb 17 '22 at 17:37

Enes Cinar

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Hey Enes, could you maybe elaborate a bit on the answer? – Rick Rackow Feb 22 '22 at 15:53
hi sorry for the late response. You can send an integer value to the `CreateMany()` function. it would return you that amount of instance you want. – Enes Cinar Jun 08 '22 at 17:27

Francois Borgies · Answer 7 · 2022-04-15T13:35:10.567

0

You can use AutoFixture.CreateMany(n) where n is the numbers of items you want :

var persons = Fixture.CreateMany<Person>(100).ToList();
//persons.Count() = 100

You can also configure AutoFixture with pre-configured numbers of items with RepeatCount :

fixture.RepeatCount = 100;
var persons = fixture.CreateMany<Person>().ToList();
//persons.Count() = 100

edited Apr 15 '22 at 13:35

answered Apr 15 '22 at 13:27

Francois Borgies

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Is it possible with Fixture to create a list of N objects?

7 Answers7